(a)
Given \( z = \frac{y – x}{y + x} \), apply the quotient rule.
Numerator: \( u = y – x \), so \(\frac{du}{dx} = \frac{dy}{dx} – 1\).
Denominator: \( v = y + x \), so \(\frac{dv}{dx} = \frac{dy}{dx} + 1\).
\(\frac{dz}{dx} = \frac{(y + x)\left( \frac{dy}{dx} – 1 \right) – (y – x)\left( \frac{dy}{dx} + 1 \right)}{(y + x)^2}\).
Expand numerator: \( (y + x)\left( \frac{dy}{dx} – 1 \right) – (y – x)\left( \frac{dy}{dx} + 1 \right) = 2x \frac{dy}{dx} – 2y \).
Thus: \(\frac{dz}{dx} = \frac{2x \frac{dy}{dx} – 2y}{(y + x)^2} = \frac{2}{(y + x)^2} \left( x \times \frac{dy}{dx} – y \right)\).
Result: \(\frac{dz}{dx} = \frac{2}{(y + x)^2} \left( x \times \frac{dy}{dx} – y \right)\) [3]

(b)
Given: \( f(x) \left( x \times \frac{dy}{dx} – y \right) = y^2 – x^2 \).
From (a): \(\frac{dz}{dx} = \frac{2}{(y + x)^2} \left( x \times \frac{dy}{dx} – y \right)\).
So: \( f(x) \frac{dz}{dx} = f(x) \cdot \frac{2}{(y + x)^2} \left( x \times \frac{dy}{dx} – y \right) = \frac{2}{(y + x)^2} (y^2 – x^2) = 2 \frac{(y – x)(y + x)}{(y + x)^2} = 2 \frac{y – x}{y + x} = 2z\).
Result: \( f(x) \frac{dz}{dx} = 2z \) [2]

(c)
Given: \(\left( x – 3 \right) \left( x \times \frac{dy}{dx} – y \right) = y^2 – x^2\), with \( x = 4 \), \( y = 5 \).
From (b): \( f(x) \frac{dz}{dx} = 2z \), where \( f(x) = x – 3 \).
So: \(\frac{dz}{dx} = \frac{2z}{x – 3}\).
Separate variables: \(\frac{dz}{z} = \frac{2}{x – 3} dx\).
Integrate: \(\ln z = 2 \ln (x – 3) + c\).
Initial condition: At \( x = 4 \), \( y = 5 \), \( z = \frac{5 – 4}{5 + 4} = \frac{1}{9} \), so \(\ln \frac{1}{9} = 2 \ln (4 – 3) + c = c \implies c = \ln \frac{1}{9}\).
Thus: \(\ln z = 2 \ln (x – 3) + \ln \frac{1}{9} = \ln \left( \frac{(x – 3)^2}{9} \right) = \ln \left( \left( \frac{x – 3}{3} \right)^2 \right)\).
Exponentiate: \( z = \left( \frac{x – 3}{3} \right)^2 \).
Since \( z = \frac{y – x}{y + x} \), the solution is: \(\frac{y – x}{y + x} = \left( \frac{x – 3}{3} \right)^2\).
Result: \(\frac{y – x}{y + x} = \left( \frac{x – 3}{3} \right)^2\) [7]